Answer is: the boiling point of the resulting solution of sucrose is 100.42°C.
m(H₂<span>O) = 15.2 g ÷ 1000 g/kg = 0.0152 kg.
</span>m(C₁₂H₂₂O₁₁<span>) = 4.27 g.
n</span>(C₁₂H₂₂O₁₁) = m(C₁₂H₂₂O₁₁) ÷ M(C₁₂H₂₂O₁₁).
n(C₁₂H₂₂O₁₁) = 4.27 g ÷ 342.3 g/mol.
n(C₁₂H₂₂O₁₁) = 0.0125 mol.
b(solution) = n(C₁₂H₂₂O₁₁) ÷ m(H₂O).
b(solution) = 0.0125 mol ÷ 0.0152 kg.
b(solution) = 0.82 m.
ΔT = b(solution) · Kb(H₂O).
ΔT = 0.82 m · 0.512°C/m.
ΔT = 0.42°C.
Tb = 100°C + 0.42°C = 100.42°C.
Answer:- 
molecules.
Solution:- The grams of tetrabromomethane are given and it asks to calculate the number of molecules.
It is a two step unit conversion problem. In the first step, grams are converted to moles on dividing the grams by molar mass.
In second step, the moles are converted to molecules on multiplying by Avogadro number.
Molar mass of 
= 12+4(79.9) = 331.6 g per mol
let's make the set up using dimensional analysis:
= molecules
So, there will be molecules in 250 grams of .
Answer: Non polar solvents
Explanation:
Since with increasing the size of alkyl group hydrophobic nature increases and solubility in polar solvents decreases .
Hence Carboxylic acids with more than 10 carbon atoms, solubility is more in non polar solvents.
2: <span>Volume V = a*b*c = 6.0*3.0*3.0 = 54.0 cm^3 density ρ = mass/volume = 146/54 = 2.70 g/cm^3
3: Volume = (27.8 -21.2) cm^3
mass = 22.4 g
density = 22.4/(27.8-21.2) g/cm^3
</span>
Answer:
Explanation:
It will be better to use solvents that are lighter than water, because their density has an influence on the miscibility . This will give you a better separation during extraction.
